Mixing
Characterization of a rotund space
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A normed linear space $X$ is said to be rotund if for all $x,yin X$ with $|x|=1=|y|$, $|x+y|<2$.
I want to prove that a normed linear space $X$ is rotund iff the function $varphi:Xto mathbb{R}$ defined by $varphi(x)=frac{1}{2}|x|^2$ for all $xin X$ is strictly convex.
Suppose $varphi$ is strictly convex. Let $x,yin X$ such that $|x|=1=|y|$. Then $|frac{x+y}{2}|^2=2varphi(frac{x}{2}+frac{y}{2})<varphi(x)+varphi(y)=frac{1}{2}+frac{1}{2}=1$. Thus $|x+y|<2$ and so $X$ is rotund. But how to prove the converse. Any hint will be thankfully appreciated.
functional-analysis convex-analysis norm normed-spaces
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
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add a comment |
$begingroup$
A normed linear space $X$ is said to be rotund if for all $x,yin X$ with $|x|=1=|y|$, $|x+y|<2$.
I want to prove that a normed linear space $X$ is rotund iff the function $varphi:Xto mathbb{R}$ defined by $varphi(x)=frac{1}{2}|x|^2$ for all $xin X$ is strictly convex.
Suppose $varphi$ is strictly convex. Let $x,yin X$ such that $|x|=1=|y|$. Then $|frac{x+y}{2}|^2=2varphi(frac{x}{2}+frac{y}{2})<varphi(x)+varphi(y)=frac{1}{2}+frac{1}{2}=1$. Thus $|x+y|<2$ and so $X$ is rotund. But how to prove the converse. Any hint will be thankfully appreciated.
functional-analysis convex-analysis norm normed-spaces
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
$endgroup$
add a comment |
$begingroup$
A normed linear space $X$ is said to be rotund if for all $x,yin X$ with $|x|=1=|y|$, $|x+y|<2$.
I want to prove that a normed linear space $X$ is rotund iff the function $varphi:Xto mathbb{R}$ defined by $varphi(x)=frac{1}{2}|x|^2$ for all $xin X$ is strictly convex.
Suppose $varphi$ is strictly convex. Let $x,yin X$ such that $|x|=1=|y|$. Then $|frac{x+y}{2}|^2=2varphi(frac{x}{2}+frac{y}{2})<varphi(x)+varphi(y)=frac{1}{2}+frac{1}{2}=1$. Thus $|x+y|<2$ and so $X$ is rotund. But how to prove the converse. Any hint will be thankfully appreciated.
functional-analysis convex-analysis norm normed-spaces
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
$endgroup$
A normed linear space $X$ is said to be rotund if for all $x,yin X$ with $|x|=1=|y|$, $|x+y|<2$.
I want to prove that a normed linear space $X$ is rotund iff the function $varphi:Xto mathbb{R}$ defined by $varphi(x)=frac{1}{2}|x|^2$ for all $xin X$ is strictly convex.
Suppose $varphi$ is strictly convex. Let $x,yin X$ such that $|x|=1=|y|$. Then $|frac{x+y}{2}|^2=2varphi(frac{x}{2}+frac{y}{2})<varphi(x)+varphi(y)=frac{1}{2}+frac{1}{2}=1$. Thus $|x+y|<2$ and so $X$ is rotund. But how to prove the converse. Any hint will be thankfully appreciated.
functional-analysis convex-analysis norm normed-spaces
functional-analysis convex-analysis norm normed-spaces
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
asked Feb 1 at 9:46
AnupamAnupam
2,5401825
2,5401825
add a comment |
add a comment |
1 Answer
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We can use a different equivalent definition of rotund given in an "An introduction to Banach space theory" by R.E. Megginson.
A normed space is rotund if for all $x,yin X$ with $|x|=|y|=1$ and all $tin (0,1)$ we have $|tx+(1-t)y|<1$. This is equivalent to your definition by Proposition 5.1.2 in this book.
Let $x,yin X$ and let $lambdain(0,1)$. We have to show that $varphi(lambda x+(1-lambda)y)<lambdavarphi(x)+(1-lambda)varphi(y)$. Let $x'=frac{x}{|x|}$, let $y'=frac{y}{|y|}$, let $c=lambda|x|+(1-lambda)|y|$ and let $mu=frac{lambda|x|}{c}in(0,1)$. Note that
$$varphi(lambda x+(1-lambda)y)=c^{2}varphi(mu x'+(1-mu)y')=frac{c^{2}}{2}|mu x'+(1-mu)y'|^{2}<frac{c^{2}}{2}\=frac{lambda^{2}|x|^{2}+2(lambda-lambda^{2})|x||y|+(1-lambda)^{2}|y|^{2}}{2}\=frac{lambda|x|^{2}+(1-lambda)|y|^{2}+(lambda^{2}-lambda)|x|^{2}+2(lambda-lambda^{2})|x||y|+(lambda^{2}-lambda)|y|^{2}}{2}=(*).$$
Since $2|x||y|leq|x|^{2}+|y|^{2}$ we find
$$(*)leqfrac{lambda|x|^{2}+(1-lambda)|y|^{2}}{2}=lambda|x|^{2}varphi(x')+(1-lambda)|y|^{2}varphi(y')=lambdavarphi(x)+(1-lambda)varphi(y).$$
So $varphi$ is strictly convex.
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
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1 Answer
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1 Answer
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$begingroup$
We can use a different equivalent definition of rotund given in an "An introduction to Banach space theory" by R.E. Megginson.
A normed space is rotund if for all $x,yin X$ with $|x|=|y|=1$ and all $tin (0,1)$ we have $|tx+(1-t)y|<1$. This is equivalent to your definition by Proposition 5.1.2 in this book.
Let $x,yin X$ and let $lambdain(0,1)$. We have to show that $varphi(lambda x+(1-lambda)y)<lambdavarphi(x)+(1-lambda)varphi(y)$. Let $x'=frac{x}{|x|}$, let $y'=frac{y}{|y|}$, let $c=lambda|x|+(1-lambda)|y|$ and let $mu=frac{lambda|x|}{c}in(0,1)$. Note that
$$varphi(lambda x+(1-lambda)y)=c^{2}varphi(mu x'+(1-mu)y')=frac{c^{2}}{2}|mu x'+(1-mu)y'|^{2}<frac{c^{2}}{2}\=frac{lambda^{2}|x|^{2}+2(lambda-lambda^{2})|x||y|+(1-lambda)^{2}|y|^{2}}{2}\=frac{lambda|x|^{2}+(1-lambda)|y|^{2}+(lambda^{2}-lambda)|x|^{2}+2(lambda-lambda^{2})|x||y|+(lambda^{2}-lambda)|y|^{2}}{2}=(*).$$
Since $2|x||y|leq|x|^{2}+|y|^{2}$ we find
$$(*)leqfrac{lambda|x|^{2}+(1-lambda)|y|^{2}}{2}=lambda|x|^{2}varphi(x')+(1-lambda)|y|^{2}varphi(y')=lambdavarphi(x)+(1-lambda)varphi(y).$$
So $varphi$ is strictly convex.
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
$endgroup$
add a comment |
$begingroup$
We can use a different equivalent definition of rotund given in an "An introduction to Banach space theory" by R.E. Megginson.
A normed space is rotund if for all $x,yin X$ with $|x|=|y|=1$ and all $tin (0,1)$ we have $|tx+(1-t)y|<1$. This is equivalent to your definition by Proposition 5.1.2 in this book.
Let $x,yin X$ and let $lambdain(0,1)$. We have to show that $varphi(lambda x+(1-lambda)y)<lambdavarphi(x)+(1-lambda)varphi(y)$. Let $x'=frac{x}{|x|}$, let $y'=frac{y}{|y|}$, let $c=lambda|x|+(1-lambda)|y|$ and let $mu=frac{lambda|x|}{c}in(0,1)$. Note that
$$varphi(lambda x+(1-lambda)y)=c^{2}varphi(mu x'+(1-mu)y')=frac{c^{2}}{2}|mu x'+(1-mu)y'|^{2}<frac{c^{2}}{2}\=frac{lambda^{2}|x|^{2}+2(lambda-lambda^{2})|x||y|+(1-lambda)^{2}|y|^{2}}{2}\=frac{lambda|x|^{2}+(1-lambda)|y|^{2}+(lambda^{2}-lambda)|x|^{2}+2(lambda-lambda^{2})|x||y|+(lambda^{2}-lambda)|y|^{2}}{2}=(*).$$
Since $2|x||y|leq|x|^{2}+|y|^{2}$ we find
$$(*)leqfrac{lambda|x|^{2}+(1-lambda)|y|^{2}}{2}=lambda|x|^{2}varphi(x')+(1-lambda)|y|^{2}varphi(y')=lambdavarphi(x)+(1-lambda)varphi(y).$$
So $varphi$ is strictly convex.
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
$endgroup$
add a comment |
$begingroup$
We can use a different equivalent definition of rotund given in an "An introduction to Banach space theory" by R.E. Megginson.
A normed space is rotund if for all $x,yin X$ with $|x|=|y|=1$ and all $tin (0,1)$ we have $|tx+(1-t)y|<1$. This is equivalent to your definition by Proposition 5.1.2 in this book.
Let $x,yin X$ and let $lambdain(0,1)$. We have to show that $varphi(lambda x+(1-lambda)y)<lambdavarphi(x)+(1-lambda)varphi(y)$. Let $x'=frac{x}{|x|}$, let $y'=frac{y}{|y|}$, let $c=lambda|x|+(1-lambda)|y|$ and let $mu=frac{lambda|x|}{c}in(0,1)$. Note that
$$varphi(lambda x+(1-lambda)y)=c^{2}varphi(mu x'+(1-mu)y')=frac{c^{2}}{2}|mu x'+(1-mu)y'|^{2}<frac{c^{2}}{2}\=frac{lambda^{2}|x|^{2}+2(lambda-lambda^{2})|x||y|+(1-lambda)^{2}|y|^{2}}{2}\=frac{lambda|x|^{2}+(1-lambda)|y|^{2}+(lambda^{2}-lambda)|x|^{2}+2(lambda-lambda^{2})|x||y|+(lambda^{2}-lambda)|y|^{2}}{2}=(*).$$
Since $2|x||y|leq|x|^{2}+|y|^{2}$ we find
$$(*)leqfrac{lambda|x|^{2}+(1-lambda)|y|^{2}}{2}=lambda|x|^{2}varphi(x')+(1-lambda)|y|^{2}varphi(y')=lambdavarphi(x)+(1-lambda)varphi(y).$$
So $varphi$ is strictly convex.
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
$endgroup$
We can use a different equivalent definition of rotund given in an "An introduction to Banach space theory" by R.E. Megginson.
A normed space is rotund if for all $x,yin X$ with $|x|=|y|=1$ and all $tin (0,1)$ we have $|tx+(1-t)y|<1$. This is equivalent to your definition by Proposition 5.1.2 in this book.
Let $x,yin X$ and let $lambdain(0,1)$. We have to show that $varphi(lambda x+(1-lambda)y)<lambdavarphi(x)+(1-lambda)varphi(y)$. Let $x'=frac{x}{|x|}$, let $y'=frac{y}{|y|}$, let $c=lambda|x|+(1-lambda)|y|$ and let $mu=frac{lambda|x|}{c}in(0,1)$. Note that
$$varphi(lambda x+(1-lambda)y)=c^{2}varphi(mu x'+(1-mu)y')=frac{c^{2}}{2}|mu x'+(1-mu)y'|^{2}<frac{c^{2}}{2}\=frac{lambda^{2}|x|^{2}+2(lambda-lambda^{2})|x||y|+(1-lambda)^{2}|y|^{2}}{2}\=frac{lambda|x|^{2}+(1-lambda)|y|^{2}+(lambda^{2}-lambda)|x|^{2}+2(lambda-lambda^{2})|x||y|+(lambda^{2}-lambda)|y|^{2}}{2}=(*).$$
Since $2|x||y|leq|x|^{2}+|y|^{2}$ we find
$$(*)leqfrac{lambda|x|^{2}+(1-lambda)|y|^{2}}{2}=lambda|x|^{2}varphi(x')+(1-lambda)|y|^{2}varphi(y')=lambdavarphi(x)+(1-lambda)varphi(y).$$
So $varphi$ is strictly convex.
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
answered Feb 1 at 10:53
Floris ClaassensFloris Claassens
1,45429
1,45429
add a comment |
add a comment |
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